【发布时间】：2017-12-02 02:34:52
【问题描述】：
在我的应用中出现此错误
W/System.err: org.json.JSONException: java.lang.String 类型的值 br 无法转换为 JSONObject
这是我的 RegisterActivity.java
public class RegisterActivity extends AppCompatActivity {
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_register);
final EditText etName=(EditText) findViewById(R.id.etname);
final EditText etUsername=(EditText) findViewById(R.id.etusername);
final EditText etPassword=(EditText) findViewById(R.id.etpassword);
final EditText etCPassword=(EditText) findViewById(R.id.etcpassword);
final EditText etEmail=(EditText) findViewById(R.id.etemail);
final EditText etContact=(EditText) findViewById(R.id.etcontact);
final EditText etAge=(EditText) findViewById(R.id.etage);
final Button btnRegister=(Button) findViewById(R.id.btnregister);
btnRegister.setOnClickListener(new View.OnClickListener() {
@Override
public void onClick(View v) {
final String name=etName.getText().toString();
final String username=etUsername.getText().toString();
final String password=etPassword.getText().toString();
final String email=etEmail.getText().toString();
final int age=Integer.parseInt(etAge.getText().toString());
final int contact=Integer.parseInt(etContact.getText().toString());
Response.Listener<String> responseListener = new Response.Listener<String>(){
@Override
public void onResponse(String response) {
try {
JSONObject jsonResponse= new JSONObject(response);
boolean success = jsonResponse.getBoolean("success");
if(success){
Intent intent = new Intent(RegisterActivity.this,LoginActivity.class);
RegisterActivity.this.startActivity(intent);
}else{
AlertDialog.Builder builder= new AlertDialog.Builder(RegisterActivity.this);
builder.setMessage("Register Failed")
.setNegativeButton("Retry",null)
.create()
.show();
}
} catch (JSONException e) {
e.printStackTrace();
AlertDialog.Builder build= new AlertDialog.Builder(RegisterActivity.this);
build.setMessage("JSON Failed")
.setNegativeButton("Retry",null)
.create()
.show();
}
}
};
RegisterRequest registerRequest = new RegisterRequest(name,username,password,email,contact,age,responseListener);
RequestQueue queue = Volley.newRequestQueue(RegisterActivity.this);
queue.add(registerRequest);
}
});
}
}
这是 RegistryRequest.java
public class RegisterRequest extends StringRequest{
private static final String REGISTER_REQUEST_URL="https://smplycode.000webhostapp.com/Registerft.php";
private Map<String, String> params;
public RegisterRequest(String name,String username,String password,String email,int contact, int age, Response.Listener<String> listener) {
super(Method.POST, REGISTER_REQUEST_URL, listener, null);
params = new HashMap<>();
params.put("name",name);
params.put("username",username);
params.put("password",password);
params.put("email",email);
params.put("contact",contact+"");
params.put("age",age+"");
}
@Override
public Map<String, String> getParams() {return params;}
}
这是我的 registerft.php 文件
<?php
$con = mysqli_connect("localhost", "id20401_mydbname", "mydbpaord", "id2041_mydme");
$name = $_POST["name"];
$username = $_POST["username"];
$password = $_POST["password"];
$email = $_POST["email"];
$contact = $_POST["contact"];
$age = $_POST["age"];
$statement = mysqli_prepare($con, "INSERT INTO usersft (name, username, password, email, contact, age) VALUES (?, ?, ?, ?, ?, ?)");
mysqli_stmt_bind_param($statement, "ssssii", $name, $username,$password,$email,$contact, $age);
mysqli_stmt_execute($statement);
$response = array();
$response["success"] = true;
echo json_encode($response);
?>
我在 youtube 上观看了 this 教程，这段代码正在处理该教程。
Getting this when printing response string as an alert
如果有人想提供帮助，请给我您的电子邮件 ID 我正在做一个项目我想为新手制作一个关于 C++ 学习的应用程序:)
【问题讨论】：
-
服务器响应是什么？
-
最重要的部分是知道你试图解析的 JSON 文本，那么如何弄清楚它是什么，正如@an_droid_dev 所说，并展示给我们。或者，也许一旦你看到它，你就会亲眼看到哪里出了问题，而这个问题是没有实际意义的。
-
请告诉我在哪里可以看到我正在使用 000.webhost.com 进行托管
-
查看答案我收到此错误（详细）
-
i.stack.imgur.com/naT1E.png 这是我打印响应字符串时得到的结果